tag:blogger.com,1999:blog-3722233.post1083432864829188275..comments2024-07-08T08:20:09.315-05:00Comments on Computational Complexity: Double Digit DelightsLance Fortnowhttp://www.blogger.com/profile/06752030912874378610noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-3722233.post-73909141844493490232024-06-06T13:13:54.710-05:002024-06-06T13:13:54.710-05:00For 3 digit numbers each digit appears in the sum ...For 3 digit numbers each digit appears in the sum 4 times, twice in the 10s place and twice in the 1s place so the number must be divisible by 22. Writing the number as abc, we have 22*(a+b+c) = 100*a + 10*b + c, which reduces to 26*a = 4*b + 7*c, and c must be even. Since b and c can be at most 9 the RHS can be at most 99 so a must be 1, 2, or 3. If a=1, then c can be at most 3 so must be 2 and we recover 132. If a=2, then c can be at most 7 and must be divisible by 4 so must be 4 and we recover 264. If a=3, then c must be at least 6 and cannot be divisible by 4 so must be 6 and we recover 396.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-79996997019198432972024-06-03T11:30:00.585-05:002024-06-03T11:30:00.585-05:00interesting.
I think generally number theorists ...interesting. <br /><br />I think generally number theorists seldom deal with digits because base 10 is kind of arbitrary after all.<br /><br />There is good amount of number theory around sum over a function applied to divisors of a number. <br /><br />I wonder if the topic of sum of a function over digits of a number in a base m is studied.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-24606018578700870952024-06-03T03:37:44.537-05:002024-06-03T03:37:44.537-05:00@B., you are working with a different definition t...@B., you are working with a different definition than Lance.<br />For your candidate 10545, your eligible summands are apparently,<br />[14,15,41,45,51,54,55,104,105,140,145,150,154,155,401,405,410,415,450,451,455,501,504,505,510,514,515,540,541,545,550,551,554] --> 10545, but this is not what this is about for 3-idempotent numbers unless you assume a leading zero for the 2 digit numbers.<br />However, for 35964 we have summands:<br />[345,346,349,354,356,359,364,365,369,394,395,396,435,436,439,453,456,459,463,465,469,493,495,496,534,536,539,543,546,549,563,564,569,593,594,596,634,635,639,643,645,649,653,654,659,693,694,695,934,935,936,943,945,946,953,954,956,963,964,965].Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-89646075604436543582024-05-29T07:36:05.040-05:002024-05-29T07:36:05.040-05:00A related (more natural?) definition would be the ...A related (more natural?) definition would be the sum of all *distinct* 2-digit numbers. Of course, 11, 22, … would be 2-idempotent for this new definition and boring. But are there others? <br /><br />The argument for ≥4-digit numbers carries on. Of course the argument for 2-digit numbers doesn't, but proves that the only 2-idempotent 2-digit integers are 11k for k = 1, …, 9. And for 3-digit numbers, a computer search shows that again the only possibilities are 132*k for k=1,2,3.<br /><br />And for 3-idempotent numbers? You get (of course) 111*k (1 ≤ k ≤ 9), but also 8991 and 10545 (in addition to 35964). <br /><br />OK, I am willing to play: the 4-idempotent numbers are 1111*k (1≤k≤9), and 255530. But no 4-idempotent number with the original definition!B.noreply@blogger.com